Question

A copper ore contains 30 wt.% chalcopyrite (\(CuFeS_2\)) and the remaining gangue material. Assuming no copper is present in the gangue material, the amount of copper in the ore (rounded off to one decimal place) is ______________ wt.%.
Given: Atomic weights of Fe, Cu and S are 56, 63.5, and 32 g/mol, respectively.

Hint:

Find the mass fraction of Cu inside CuFeS2 using the atomic weights, then scale it by the 30 wt.% chalcopyrite content of the ore.

Answer 1

Correct Answer: 10.3

Step 1: Find the molar mass of chalcopyrite.
Chalcopyrite has the formula \(CuFeS_2\), one atom each of copper and iron, and two atoms of sulfur. Using the given atomic weights,
\[ M_{CuFeS_2} = 63.5 + 56 + 2(32) = 63.5+56+64 = 183.5 \text{ g/mol} \]

Step 2: Find the fraction of copper by mass inside chalcopyrite.
Only the copper atom in the formula unit is the metal we want, so its mass fraction inside the mineral is its own weight divided by the whole formula weight.
\[ w_{Cu \text{ in } CuFeS_2} = \frac{63.5}{183.5} = 0.3460 \]
This is about 34.6% of the chalcopyrite by mass.

Step 3: Scale down to the whole ore.
The ore is only 30 wt.% chalcopyrite, and the rest is gangue with zero copper in it. So the copper content of the whole ore is the copper fraction inside chalcopyrite multiplied by the amount of chalcopyrite present.
\[ w_{Cu \text{ in ore}} = 0.3460 \times 30\% = 10.38\% \]

Final Answer:
The ore carries about 10.4 wt.% copper, which lies inside the accepted band of 10.3 to 10.5 wt.%.
\[ \boxed{w_{Cu} \approx 10.4\text{ wt.\%}} \]